3.24.2 \(\int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)^{3/2}} \, dx\)

Optimal. Leaf size=108 \[ -\frac {22 (1-2 x)^{3/2}}{5 \sqrt {5 x+3}}-\frac {128}{75} \sqrt {5 x+3} \sqrt {1-2 x}+\frac {338}{225} \sqrt {\frac {2}{5}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )+\frac {98}{9} \sqrt {7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right ) \]

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Rubi [A]  time = 0.04, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {98, 154, 157, 54, 216, 93, 204} \begin {gather*} -\frac {22 (1-2 x)^{3/2}}{5 \sqrt {5 x+3}}-\frac {128}{75} \sqrt {5 x+3} \sqrt {1-2 x}+\frac {338}{225} \sqrt {\frac {2}{5}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )+\frac {98}{9} \sqrt {7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - 2*x)^(5/2)/((2 + 3*x)*(3 + 5*x)^(3/2)),x]

[Out]

(-22*(1 - 2*x)^(3/2))/(5*Sqrt[3 + 5*x]) - (128*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/75 + (338*Sqrt[2/5]*ArcSin[Sqrt[2/
11]*Sqrt[3 + 5*x]])/225 + (98*Sqrt[7]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/9

Rule 54

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[2/Sqrt[b], Subst[Int[1/Sqrt[b*c -
a*d + d*x^2], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[b*c - a*d, 0] && GtQ[b, 0]

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] + Dist[1/(b*(b*e - a*
f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 154

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegersQ[2*m, 2
*n, 2*p]

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin {align*} \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)^{3/2}} \, dx &=-\frac {22 (1-2 x)^{3/2}}{5 \sqrt {3+5 x}}-\frac {2}{5} \int \frac {\sqrt {1-2 x} \left (\frac {167}{2}+64 x\right )}{(2+3 x) \sqrt {3+5 x}} \, dx\\ &=-\frac {22 (1-2 x)^{3/2}}{5 \sqrt {3+5 x}}-\frac {128}{75} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {2}{75} \int \frac {\frac {2633}{2}-169 x}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx\\ &=-\frac {22 (1-2 x)^{3/2}}{5 \sqrt {3+5 x}}-\frac {128}{75} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {338}{225} \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx-\frac {343}{9} \int \frac {1}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx\\ &=-\frac {22 (1-2 x)^{3/2}}{5 \sqrt {3+5 x}}-\frac {128}{75} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {686}{9} \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,\frac {\sqrt {1-2 x}}{\sqrt {3+5 x}}\right )+\frac {676 \operatorname {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )}{225 \sqrt {5}}\\ &=-\frac {22 (1-2 x)^{3/2}}{5 \sqrt {3+5 x}}-\frac {128}{75} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {338}{225} \sqrt {\frac {2}{5}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )+\frac {98}{9} \sqrt {7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.82, size = 140, normalized size = 1.30 \begin {gather*} \frac {2 \left (6 \sqrt {22} (1-2 x)^{5/2} \, _2F_1\left (\frac {3}{2},\frac {5}{2};\frac {7}{2};-\frac {5}{11} (2 x-1)\right )+847 \left (-\frac {33 \sqrt {1-2 x}}{\sqrt {5 x+3}}+35 \sqrt {7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )+\frac {2 \sqrt {\frac {2}{5}} (2 x-1) \sinh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {2 x-1}\right )}{\sqrt {-(1-2 x)^2}}\right )\right )}{5445} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - 2*x)^(5/2)/((2 + 3*x)*(3 + 5*x)^(3/2)),x]

[Out]

(2*(847*((-33*Sqrt[1 - 2*x])/Sqrt[3 + 5*x] + (2*Sqrt[2/5]*(-1 + 2*x)*ArcSinh[Sqrt[5/11]*Sqrt[-1 + 2*x]])/Sqrt[
-(1 - 2*x)^2] + 35*Sqrt[7]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])]) + 6*Sqrt[22]*(1 - 2*x)^(5/2)*Hyperge
ometric2F1[3/2, 5/2, 7/2, (-5*(-1 + 2*x))/11]))/5445

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IntegrateAlgebraic [A]  time = 0.19, size = 129, normalized size = 1.19 \begin {gather*} -\frac {22 \sqrt {1-2 x} \left (\frac {165 (1-2 x)}{5 x+3}+64\right )}{75 \sqrt {5 x+3} \left (\frac {5 (1-2 x)}{5 x+3}+2\right )}-\frac {338}{225} \sqrt {\frac {2}{5}} \tan ^{-1}\left (\frac {\sqrt {\frac {5}{2}} \sqrt {1-2 x}}{\sqrt {5 x+3}}\right )+\frac {98}{9} \sqrt {7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(1 - 2*x)^(5/2)/((2 + 3*x)*(3 + 5*x)^(3/2)),x]

[Out]

(-22*Sqrt[1 - 2*x]*(64 + (165*(1 - 2*x))/(3 + 5*x)))/(75*Sqrt[3 + 5*x]*(2 + (5*(1 - 2*x))/(3 + 5*x))) - (338*S
qrt[2/5]*ArcTan[(Sqrt[5/2]*Sqrt[1 - 2*x])/Sqrt[3 + 5*x]])/225 + (98*Sqrt[7]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt
[3 + 5*x])])/9

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fricas [A]  time = 1.41, size = 127, normalized size = 1.18 \begin {gather*} -\frac {169 \, \sqrt {5} \sqrt {2} {\left (5 \, x + 3\right )} \arctan \left (\frac {\sqrt {5} \sqrt {2} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 6125 \, \sqrt {7} {\left (5 \, x + 3\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 30 \, {\left (10 \, x - 357\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{1125 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)^(5/2)/(2+3*x)/(3+5*x)^(3/2),x, algorithm="fricas")

[Out]

-1/1125*(169*sqrt(5)*sqrt(2)*(5*x + 3)*arctan(1/20*sqrt(5)*sqrt(2)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10
*x^2 + x - 3)) - 6125*sqrt(7)*(5*x + 3)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 +
x - 3)) - 30*(10*x - 357)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(5*x + 3)

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giac [B]  time = 1.86, size = 219, normalized size = 2.03 \begin {gather*} -\frac {49}{90} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} + \frac {169}{1125} \, \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{4 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} + \frac {4}{375} \, \sqrt {5} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - \frac {121}{250} \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)^(5/2)/(2+3*x)/(3+5*x)^(3/2),x, algorithm="giac")

[Out]

-49/90*sqrt(70)*sqrt(10)*(pi + 2*arctan(-1/140*sqrt(70)*sqrt(5*x + 3)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/
(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))) + 169/1125*sqrt(10)*(pi + 2*arctan(-1/4*sqrt(5*x + 3)*((
sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))) + 4/375*sqrt(5)*sq
rt(5*x + 3)*sqrt(-10*x + 5) - 121/250*sqrt(10)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 4*sqrt(5*
x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))

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maple [A]  time = 0.01, size = 139, normalized size = 1.29 \begin {gather*} \frac {\left (845 \sqrt {10}\, x \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-30625 \sqrt {7}\, x \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+300 \sqrt {-10 x^{2}-x +3}\, x +507 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-18375 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-10710 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {-2 x +1}}{1125 \sqrt {-10 x^{2}-x +3}\, \sqrt {5 x +3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x+1)^(5/2)/(3*x+2)/(5*x+3)^(3/2),x)

[Out]

1/1125*(845*10^(1/2)*x*arcsin(20/11*x+1/11)-30625*7^(1/2)*x*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))
+507*10^(1/2)*arcsin(20/11*x+1/11)-18375*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+300*(-10*x
^2-x+3)^(1/2)*x-10710*(-10*x^2-x+3)^(1/2))*(-2*x+1)^(1/2)/(-10*x^2-x+3)^(1/2)/(5*x+3)^(1/2)

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maxima [A]  time = 1.27, size = 86, normalized size = 0.80 \begin {gather*} -\frac {8 \, x^{2}}{15 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {169}{1125} \, \sqrt {10} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) - \frac {49}{9} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {1448 \, x}{75 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {238}{25 \, \sqrt {-10 \, x^{2} - x + 3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)^(5/2)/(2+3*x)/(3+5*x)^(3/2),x, algorithm="maxima")

[Out]

-8/15*x^2/sqrt(-10*x^2 - x + 3) + 169/1125*sqrt(10)*arcsin(20/11*x + 1/11) - 49/9*sqrt(7)*arcsin(37/11*x/abs(3
*x + 2) + 20/11/abs(3*x + 2)) + 1448/75*x/sqrt(-10*x^2 - x + 3) - 238/25/sqrt(-10*x^2 - x + 3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (1-2\,x\right )}^{5/2}}{\left (3\,x+2\right )\,{\left (5\,x+3\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - 2*x)^(5/2)/((3*x + 2)*(5*x + 3)^(3/2)),x)

[Out]

int((1 - 2*x)^(5/2)/((3*x + 2)*(5*x + 3)^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (1 - 2 x\right )^{\frac {5}{2}}}{\left (3 x + 2\right ) \left (5 x + 3\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)**(5/2)/(2+3*x)/(3+5*x)**(3/2),x)

[Out]

Integral((1 - 2*x)**(5/2)/((3*x + 2)*(5*x + 3)**(3/2)), x)

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